$f$ is defined on $C[-1, 1]$ by $$f(x)=\int_{-1}^0 x(t) dt - \int_0^1 x(t) dt.$$
I can show that $\|f\| \le 2$. I don't know how to show $\|f\| \ge 2$.
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1do you have the sup norm on $C[-1,1]$? – Giovanni Sep 15 '15 at 03:57
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@Giovanni No, max norm. I'm assuming. This is not given in the problem. – charlie Sep 15 '15 at 04:00
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3Max and sup norm are the same on compact sets (since continuous functions always achieve their maximum). – Cameron Williams Sep 15 '15 at 04:04
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@CameronWilliams You're right, thanks. – charlie Sep 15 '15 at 04:10
1 Answers
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Of course if you can plug in
$$x(t) = \begin{cases} 1 & \text{if } t\in [-1, 0] \\ -1 & \text{if } t\in (0,1]\end{cases}$$
Then this function satifies $f(x) = 2$. But this $x$ is not continuous. However you can approximate this by $x_n \in C([-1,1])$, where
$$x_n (t) = \begin{cases} 1 & \text{if } t\in [-1, -\frac 1n)\\ -1 & \text{if } t\in (\frac 1n, 1]\\ - nx & \text{if } t\in [-\frac 1n, \frac 1n] \end{cases}$$
Then $\|x_n\| = 1$ and $$f(x_n) = 1- \frac 1n + \frac 1{2n} + 1- \frac 1n + \frac 1{2n} = 2 - \frac{1}{n}$$
Thus $\|f\| \ge 2- \frac{1}{n}$ for all $n$.